BASIC INSTRUMENTATION ELECTRICITY. 1.5V. Voltage and Current. R. R. I. Q=P/R. I=V/R. Liquid flow. Electrical current. Resistance. Every substance has resistance Conductor is substance having low resistance Isolator is substance having high resistance
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I=16 mA
PT
V=24 V
length= 500 m
What is the voltage across the PT
R
Vm
AC VOLTAGE AND CURRENTv(t)
0
ωt
2π
v(t) = Vmcos(ωt + )
Frequency , f = 50 Hz/ 60 Hz
T = 1/f = 1/50 = 0.02 s
ω = 2πf
is the phase angle
VAC
R
Vm
Im
ωt
i(t)
ωt
V and I in resistor are in phase
AC VOLTAGE and CURRENT in RESISTORv(t) = Vmcos ωt
i(t) = v(t)/R
= Vm (cos ωt)/R
= Imcosωt
Im = Vm/R
ωt
2π
i(t)
AC VOLTAGE, CURRENT and POWER in RESISTORv(t) = Vmcosωt
i(t) = Imcosωt
p(t) =v(t) i(t)
= VmImcos2ωt
= VmIm {1+cos(2ωt )}/2
p(t)
Vm
v(t)
i(t)
ωt
0
2π
π/2
VOLTAGE AND CURRENT IN CAPACITORv(t) = Vmcos(ωt +)
VAC
C
The current lead the voltage
C
p(t)
POWER IN CAPACITORv(t)
i(t)
2π
v(t) = Vmcos(ωt +)
+
+
p(t) = v(t) i(t)
= VmImcos(ωt + )sin(ωt + )


p(t)
POWER IN INDUCTORv(t)
i(t)
2π
v(t) = Vmsin(ωt +)
+
+

p(t) = v(t) i(t)
= VmImcos(ωt + )sin(ωt + )

R
Vm/√2
v(t)
Im /√2
i(t)
ωt
Example for V and I phasor in resistorv(t) = Vmcos(ωt + )
i(t) = Vm/Rcos(ωt + )
Vm/√2
Example for V and I phasor in capacitorIm/√2
Vm/√2
We can set the angle arbitrarily. Usually we set the voltage is set to be zero phase abritrary
v(t) = Vmcos ωt
Im/√2
Vm/√2
Im/√2
Example for V and I phasor in inductorv(t)
i(t)
Here we can set the voltage to be zero phase, then the phase of current will be
Z = V/I
V = IZ
1mF

Impedance Example:f = 50Hz
Find ZC
Answer:
Zc = 1/jwC
w=2pf =2 × 3.14 × 50 = 314 rad/s
Zc = 1/jwC=1/(j × 314 × 106)
Zc = j3184.71
R = 1K2
C = 15 mF
w = 314
ZT = ?
Answer:
Zc = 1/jwC=1/(j × 314 × 15 × 106)
Zc = j212.31
ZT = 1200 – j212.31
R = 1K2
ZT = ?
L = 5 mH
w = 314
Answer:
ZL = jwL=j × 314 × 5 × 103
ZL = j1.57
ZT = 1200 + j1.57
C = 15 mF
w = 314
Zc = j212.31
ZT = ?
L = 5 mH
w = 314
ZL = j1.57
Impedance in series exampleAnswer:
ZT = 1200 – j212.31 + j1.57
= 1200 –j210.74
Generally impedance consist of:
The real part which is called Resistance, and
The imaginary part which is called reactance
Z = R + jX
impedance
Resistance
reactance
ZR = 20kW= 20kW 0
ZC = 1/j (377 1mF) = 2.65kW 90
Complex power is defined as
S = VI*
The unit of complex power is Volt Ampere (VA)
S= VI* = I2Z = I2(R+jX) = I2R+jI2X
= I2Z cos +jI2Z sin
S = VI cos +jVI sin = P + jQ
Sis called apparent power and the unit is va
Pis called active power and the unit is watt and
Qis called reactive power and the unit is var
The term operational amplifier or "opamp" refers to a class of highgain DC coupled amplifiers with two inputs and a single output. The modern integrated circuit version is typified by the famous 741 opamp. Some of the general characteristics of the IC version are:
For an ideal opamp, the noninverting amplifier gain is given by
The voltage follower with an ideal op amp gives simply
But this turns out to be a very useful service, because the input impedance of the op amp is very high, giving effective isolation of the output from the signal source. You draw very little power from the signal source, avoiding "loading" effects. This circuit is a useful first stage.
The voltage follower is often used for the construction of buffer for logic circuits.
A circuit for converting small current signals (>0.01 microamps) to a more easily measured proportional voltage.
so the output voltage is given by the expression above.
The current output through the load resistor is proportional to the input voltage